Report 5
Intermediate Engineering Analysis
Section 7566
Team 11
Due date: March 30, 2012.
Solved by: Andrea Vargas
Problem Statement
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Part 1:Determine whether the following are linearly independent using the Wronskian
Part 2: Determine whether the following are linearly independent using the Gramian
Solution
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Using the Wronskian we check for linear independence.
We know from (1) and (2) in 7-35 that if W = det [ f g f ′ g ′ ] = f g ′ − g f ′ ≠ 0 {\displaystyle W=\det {\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}=fg'-gf'\neq 0\!} Then the functions are linearly independent.
Part 1.1
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f ( x ) = x 2 {\displaystyle f(x)=x^{2}\!} g ( x ) = x 4 {\displaystyle g(x)=x^{4}\!}
Taking the derivatives of each function:
f ′ ( x ) = 2 x {\displaystyle f'(x)=2x\!} g ′ ( x ) = 4 x 3 {\displaystyle g'(x)=4x^{3}\!}
W = [ x 2 x 4 2 x 4 x 3 ] = 4 x 5 − 2 x 5 = 2 x 5 ≠ 0 {\displaystyle W={\begin{bmatrix}x^{2}&x^{4}\\2x&4x^{3}\end{bmatrix}}=4x^{5}-2x^{5}=2x^{5}\neq 0\!}
∴ {\displaystyle \therefore \!} f(x) and g(x) are linearly independent
Part 1.2
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f ( x ) = cos ( x ) {\displaystyle f(x)=\cos(x)\!} g ( x ) = sin ( 3 x ) {\displaystyle g(x)=\sin(3x)\!}
Taking the derivatives of each function:
f ′ ( x ) = − sin ( x ) {\displaystyle f'(x)=-\sin(x)\!} g ′ ( x ) = 3 cos ( 3 x ) {\displaystyle g'(x)=3\cos(3x)\!}
W = [ cos ( x ) sin ( 3 x ) − sin ( x ) 3 cos ( 3 x ) ] = 3 cos ( x ) cos ( 3 x ) + sin ( 3 x ) sin ( x ) =≠ 0 {\displaystyle W={\begin{bmatrix}\cos(x)&\sin(3x)\\-\sin(x)&3\cos(3x)\end{bmatrix}}=3\cos(x)\cos(3x)+\sin(3x)\sin(x)=\neq 0\!}
∴ {\displaystyle \therefore \!} f(x) and g(x) are linearly independent
Using the Gramian we check for linear independence.
We know from the notes in (1) 7-34 that:⟨ f , g ⟩ := ∫ a b = f ( x ) g ( x ) {\displaystyle \langle f,g\rangle :=\int _{a}^{b}=f(x)g(x)\!}
and that the Gramian is defined as:
Γ ( f , g ) = d e t [ ⟨ f , f ⟩ ⟨ f , g ⟩ ⟨ g , f ⟩ ⟨ g , g ⟩ ] {\displaystyle \Gamma (f,g)=det{\begin{bmatrix}\langle f,f\rangle &\langle f,g\rangle \\\langle g,f\rangle &\langle g,g\rangle \end{bmatrix}}\!}
Then f,g are linearly independent if Γ ( f , g ) ≠ 0 {\displaystyle \Gamma (f,g)\neq 0\!}
Part 2.1
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f ( x ) = x 2 {\displaystyle f(x)=x^{2}\!} g ( x ) = x 4 {\displaystyle g(x)=x^{4}\!}
Taking scalar products:⟨ f , f ⟩ := ∫ − 1 1 = f ( x ) f ( x ) = x 2 ∗ x 2 = ∫ − 1 1 = x 4 = [ x 5 5 ] − 1 1 = 2 5 {\displaystyle \langle f,f\rangle :=\int _{-1}^{1}=f(x)f(x)=x^{2}*x^{2}=\int _{-1}^{1}=x^{4}=\left[{\frac {x^{5}}{5}}\right]_{-1}^{1}={\frac {2}{5}}\!} ⟨ f , g ⟩ = ⟨ g , f ⟩ := ∫ − 1 1 = f ( x ) g ( x ) = x 2 ∗ x 4 = ∫ − 1 1 = x 6 = [ x 7 7 ] − 1 1 = 2 7 {\displaystyle \langle f,g\rangle =\langle g,f\rangle :=\int _{-1}^{1}=f(x)g(x)=x^{2}*x^{4}=\int _{-1}^{1}=x^{6}=\left[{\frac {x^{7}}{7}}\right]_{-1}^{1}={\frac {2}{7}}\!} ⟨ g , g ⟩ := ∫ − 1 1 = g ( x ) g ( x ) = x 4 ∗ x 4 = ∫ − 1 1 = x 8 = [ x 9 9 ] − 1 1 = 2 9 {\displaystyle \langle g,g\rangle :=\int _{-1}^{1}=g(x)g(x)=x^{4}*x^{4}=\int _{-1}^{1}=x^{8}=\left[{\frac {x^{9}}{9}}\right]_{-1}^{1}={\frac {2}{9}}\!}
Γ ( f , g ) = det [ 2 5 2 7 2 7 2 9 ] = 0.08888 − 0.0816326531 ≠ 0 {\displaystyle \Gamma (f,g)=\det {\begin{bmatrix}{\frac {2}{5}}&{\frac {2}{7}}\\&\\{\frac {2}{7}}&{\frac {2}{9}}\end{bmatrix}}=0.08888-0.0816326531\neq 0\!}
∴ {\displaystyle \therefore \!} f(x) and g(x) are linearly independent
Part 2.2
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f ( x ) = cos ( x ) {\displaystyle f(x)=\cos(x)\!} g ( x ) = sin ( 3 x ) {\displaystyle g(x)=\sin(3x)\!}
Taking scalar products:⟨ f , f ⟩ := ∫ − 1 1 = f ( x ) f ( x ) = ∫ − 1 1 = cos 2 ( x ) d x {\displaystyle \langle f,f\rangle :=\int _{-1}^{1}=f(x)f(x)=\int _{-1}^{1}=\cos ^{2}(x)dx\!}
We can use the trig identity for power reduction cos 2 ( x ) = 1 2 cos ( 2 x ) + 1 2 {\displaystyle \cos ^{2}(x)={\frac {1}{2}}\cos(2x)+{\frac {1}{2}}\!}
Then we have, ∫ − 1 1 1 2 cos ( 2 x ) + 1 2 d x = [ 1 4 sin ( 2 x ) + 1 2 x ] − 1 1 {\displaystyle \int _{-1}^{1}{\frac {1}{2}}\cos(2x)+{\frac {1}{2}}dx=\left[{\frac {1}{4}}\sin(2x)+{\frac {1}{2}}x\right]_{-1}^{1}\!} = 0.7273243567 − ( − 0.7273243567 ) = 1.454648713 {\displaystyle =0.7273243567-(-0.7273243567)=1.454648713\!}
⟨ f , g ⟩ = ⟨ g , f ⟩ := ∫ − 1 1 f ( x ) g ( x ) d x {\displaystyle \langle f,g\rangle =\langle g,f\rangle :=\int _{-1}^{1}f(x)g(x)dx\!} = ∫ − 1 1 sin ( 3 x ) cos ( x ) d x = [ 1 8 ( − 2 cos ( 2 x ) − cos ( 4 x ) ) ] − 1 1 {\displaystyle =\int _{-1}^{1}\sin(3x)\cos(x)dx=\left[{\frac {1}{8}}(-2\cos(2x)-\cos(4x))\right]_{-1}^{1}\!} = 0.185742 − 0.185742 = 0 {\displaystyle =0.185742-0.185742=0\!}
⟨ g , g ⟩ := ∫ − 1 1 g ( x ) g ( x ) d x = ∫ − 1 1 sin ( 3 x ) sin ( 3 x ) d x {\displaystyle \langle g,g\rangle :=\int _{-1}^{1}g(x)g(x)dx=\int _{-1}^{1}\sin(3x)\sin(3x)dx\!} = 1 2 ∫ − 1 1 1 − cos ( 6 x ) d x = [ 1 2 ( x − 1 6 sin ( 6 x ) ) ] − 1 1 {\displaystyle ={\frac {1}{2}}\int _{-1}^{1}1-\cos(6x)dx=\left[{\frac {1}{2}}(x-{\frac {1}{6}}\sin(6x))\right]_{-1}^{1}\!} = 0.523285 + 0.523285 = 1.04656925 {\displaystyle =0.523285+0.523285=1.04656925\!}
Γ ( f , g ) = det [ 1.454648713 0 0 1.04656925 ] = 1.522390612 ≠ 0 {\displaystyle \Gamma (f,g)=\det {\begin{bmatrix}1.454648713&0\\0&1.04656925\end{bmatrix}}=1.522390612\neq 0\!}
∴ {\displaystyle \therefore \!} f(x) and g(x) are linearly independent
Conclusion
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By both methods (the Wronskian and the Gramian) we obtain the same results.
Problem Statement
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Verify using the Gramian that the following two vectors are linearly independent.
b 1 = 2 e 1 + 7 e 2 {\displaystyle \mathbf {b_{1}} =2\mathbf {e_{1}} +7\mathbf {e_{2}} \!} b 1 = 1.5 e 1 + 3 e 2 {\displaystyle \mathbf {b_{1}} =1.5\mathbf {e_{1}} +3\mathbf {e_{2}} \!}
Solution
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Γ ( f , g ) = d e t [ ⟨ b 1 , b 1 ⟩ ⟨ b 1 , b 2 ⟩ ⟨ b 2 , b 1 ⟩ ⟨ b 2 , b 2 ⟩ ] {\displaystyle \Gamma (f,g)=det{\begin{bmatrix}\langle \mathbf {b1} ,\mathbf {b1} \rangle &\langle \mathbf {b1} ,\mathbf {b2} \rangle \\\langle \mathbf {b2} ,\mathbf {b1} \rangle &\langle \mathbf {b2} ,\mathbf {b2} \rangle \end{bmatrix}}\!}
We know from (3) 8-9 that:⟨ b 1 , b 2 ⟩ = b 1 ⋅ b 2 < b r > {\displaystyle \langle \mathbf {b1} ,\mathbf {b2} \rangle =\mathbf {b1} \cdot \mathbf {b2} <br>\!}
We obtain,⟨ b 1 , b 1 ⟩ = ( 2 ) ( 2 ) + ( 7 ) ( 7 ) = 4 + 49 = 53 {\displaystyle \langle \mathbf {b1} ,\mathbf {b1} \rangle =(2)(2)+(7)(7)=4+49=53\!} ⟨ b 1 , b 2 ⟩ = ⟨ b 2 , b 1 ⟩ = ( 1.5 ) ( 2 ) + ( 7 ) ( 3 ) = 3 + 21 = 24 {\displaystyle \langle \mathbf {b1} ,\mathbf {b2} \rangle =\langle \mathbf {b2} ,\mathbf {b1} \rangle =(1.5)(2)+(7)(3)=3+21=24\!}
⟨ b 2 , b 2 ⟩ = ( 1.5 ) ( 1.5 ) + ( 3 ) ( 3 ) = 2.25 + 9 = 11.25 {\displaystyle \langle \mathbf {b2} ,\mathbf {b2} \rangle =(1.5)(1.5)+(3)(3)=2.25+9=11.25\!}
Then,Γ ( f , g ) = d e t [ 53 24 24 11.25 ] = 596.25 − 576 = 20.25 ≠ 0 {\displaystyle \Gamma (f,g)=det{\begin{bmatrix}53&24\\24&11.25\end{bmatrix}}=596.25-576=20.25\neq 0\!}
∴ {\displaystyle \therefore \!} b_1 and b_2 are linearly independent
Problem Statement
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Show that y p ( x ) = ∑ i = 0 n y p , i ( x ) {\displaystyle y_{p}(x)=\sum _{i=0}^{n}y_{p,i}(x)\!} is indeed the overall particular solution of the L2-ODE-VC y ″ + p ( x ) y ′ + q ( x ) y = r ( x ) {\displaystyle y''+p(x)y'+q(x)y=r(x)\!} with the excitation r ( x ) = r 1 ( x ) + r 2 ( x ) + . . . + r n ( x ) = ∑ i = 0 n r i ( x ) {\displaystyle r(x)=r_{1}(x)+r_{2}(x)+...+r_{n}(x)=\sum _{i=0}^{n}r_{i}(x)\!} .
Discuss the choice of y p ( x ) {\displaystyle y_{p}(x)\!} in the above table e.g., for r ( x ) = k cos ω x {\displaystyle r(x)=k\cos \omega x\!} why would you need to have both cos ω x {\displaystyle \cos \omega x\!} and sin ω x {\displaystyle \sin \omega x\!} in y p ( x ) {\displaystyle y_{p}(x)\!} ?
Solution
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Because the ODE is a linear equation in y and its derivatives with respect to x, the superposition principle can be applied:
r 1 ( x ) {\displaystyle r_{1}(x)\!} is a specific excitation with known form of y p 1 ( x ) {\displaystyle y_{p1}(x)\!} and r 2 ( x ) {\displaystyle r_{2}(x)\!} is a specific excitation with known form of y p 2 ( x ) {\displaystyle y_{p2}(x)\!}
+ { y p 1 ″ + p ( x ) y p 1 ′ + q ( x ) y p 1 = r 1 ( x ) y p 2 ″ + p ( x ) y p 2 ′ + q ( x ) y p 2 = r 2 ( x ) {\displaystyle +{\begin{cases}&{\text{ }}y_{p1}''+p(x)y_{p1}'+q(x)y_{p1}=r_{1}(x)\\&{\text{ }}y_{p2}''+p(x)y_{p2}'+q(x)y_{p2}=r_{2}(x)\end{cases}}\!}
becomes
( y p 1 + y p 2 ) ″ + p ( x ) ( y p 1 + y p 2 ) ′ + q ( x ) ( y p 1 + y p 2 ) = r 1 ( x ) + r 2 ( x ) {\displaystyle (y_{p1}+y_{p2})''+p(x)(y_{p1}+y_{p2})'+q(x)(y_{p1}+y_{p2})=r_{1}(x)+r_{2}(x)\!}
proving that
y p ( x ) = ∑ i = 0 n y p , i ( x ) {\displaystyle y_{p}(x)=\sum _{i=0}^{n}y_{p,i}(x)\!} is indeed the overall particular solution of the L2-ODE-VC y ″ + p ( x ) y ′ + q ( x ) y = r ( x ) {\displaystyle y''+p(x)y'+q(x)y=r(x)\!} with the excitation r ( x ) = ∑ i = 0 n r i ( x ) {\displaystyle r(x)=\sum _{i=0}^{n}r_{i}(x)\!}
According to Fourier Theorem periodic functions can be represented as infinite series in terms of cosines and sines:
f ( x ) = a 0 + ∑ n = 1 ∞ [ a n cos ω x + b n sin ω x ] {\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}\cos \omega x+b_{n}\sin \omega x]\!}
where the coefficients a 0 , a n , b n {\displaystyle a_{0},a_{n},b_{n}\!} are the Fourier coefficients calculated using Euler formulas.
So even though the system is being excited by functions like r ( x ) = k cos ω x {\displaystyle r(x)=k\cos \omega x\!} the particular solution would still include both sin ω x {\displaystyle \sin \omega x\!} and cos ω x {\displaystyle \cos \omega x\!} in y p ( x ) {\displaystyle y_{p}(x)\!} because the excitation is a periodic function that can be represented as the Fourier infinite series in terms of both sin ω x {\displaystyle \sin \omega x\!} and cos ω x {\displaystyle \cos \omega x\!} times the Fourier coefficients
Problem Statement
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Show that c o s ( 7 x ) {\displaystyle cos(7x)} and s i n ( 7 x ) {\displaystyle sin(7x)} are linearly independant using the Wronskian and the Gramain (integrate over 1 period)
Solution
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f = c o s ( 7 x ) , g = s i n ( 7 x ) {\displaystyle f=cos(7x),g=sin(7x)}
One period of 7 x = π / 7 {\displaystyle 7x=\pi /7}
Wronskian of f and gW ( f , g ) = d e t [ f g f ′ g ′ ] {\displaystyle W(f,g)=det{\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}}
Plugging in values for f , f ′ , g , g ′ ; {\displaystyle f,f',g,g';} W ( f , g ) = d e t [ c o s ( 7 x ) s i n ( 7 x ) − s i n ( 7 x ) c o s ( 7 x ) ] {\displaystyle W(f,g)=det{\begin{bmatrix}cos(7x)&sin(7x)\\-sin(7x)&cos(7x)\end{bmatrix}}}
= 7 c o s 2 ( 7 x ) + 7 s i n 2 ( 7 x ) {\displaystyle =7cos^{2}(7x)+7sin^{2}(7x)} = 7 [ c o s 2 ( 7 x ) + s i n 2 ( 7 x ) ] {\displaystyle =7[cos^{2}(7x)+sin^{2}(7x)]} = 7 [ 1 ] {\displaystyle =7[1]}
They are linearly Independant using the Wronskian.
< f , g >= ∫ a b f ( x ) g ( x ) d x {\displaystyle <f,g>=\int _{a}^{b}f(x)g(x)dx} Γ ( f , g ) = d e t [ < f , f > < f , g > < g , f > < g , g > ] {\displaystyle \Gamma (f,g)=det{\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}} ∫ 0 π / 7 c o s 2 ( 7 x ) d x = π / 14 {\displaystyle \int _{0}^{\pi /7}cos^{2}(7x)dx=\pi /14} ∫ 0 π / 7 s i n 2 ( 7 x ) d x = π / 14 {\displaystyle \int _{0}^{\pi /7}sin^{2}(7x)dx=\pi /14} ∫ 0 π / 7 c o s ( 7 x ) ∗ s i n ( 7 x ) d x = 0 {\displaystyle \int _{0}^{\pi /7}cos(7x)*sin(7x)dx=0} Γ ( f , g ) = d e t [ π / 14 0 0 π / 14 ] {\displaystyle \Gamma (f,g)=det{\begin{bmatrix}\pi /14&0\\0&\pi /14\end{bmatrix}}} Γ ( f , g ) = π 2 / 49 {\displaystyle \Gamma (f,g)=\pi ^{2}/49}
They are linearly Independent using the Gramain.
Problem Statement
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Find 2 equations for the 2 unknowns M,N and solve for M,N.
Solution
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y p ( x ) = M c o s 7 x + N s i n 7 x {\displaystyle y_{p}(x)=Mcos7x+Nsin7x} y p ′ ( x ) = − M 7 s i n 7 x + N 7 c o s 7 x {\displaystyle y'_{p}(x)=-M7sin7x+N7cos7x} y p ″ ( x ) = − M 7 2 c o s 7 x − N 7 2 s i n 7 x {\displaystyle y''_{p}(x)=-M7^{2}cos7x-N7^{2}sin7x}
Plugging these values into the equation given (y ″ − 3 y ′ − 10 y = 3 c o s 7 x {\displaystyle y''-3y'-10y=3cos7x} ) yields;− M 7 2 c o s 7 x − N 7 2 s i n 7 x − 3 ( − M 7 s i n 7 x + N 7 c o s 7 x ) − 10 ( M c o s 7 x + N s i n 7 x ) = 3 c o s 7 x {\displaystyle -M7^{2}cos7x-N7^{2}sin7x-3(-M7sin7x+N7cos7x)-10(Mcos7x+Nsin7x)=3cos7x}
Simplifying and the equating the coefficients relating sin and cos results in;− 59 M − 21 N = 3 {\displaystyle -59M-21N=3} − 59 N + 21 M = 0 {\displaystyle -59N+21M=0}
Solving for M and N results in;
M = − 177 / 3922 , N = − 63 / 3922 {\displaystyle M=-177/3922,N=-63/3922}
Problem Statement
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Find the overall solution y ( x ) {\displaystyle y(x)} that corresponds to the initial conditions y ( 0 ) = 1 , y ′ ( 0 ) = 0 {\displaystyle y(0)=1,y'(0)=0} . Plot over three periods.
Solution
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From before, one period = π / 7 {\displaystyle =\pi /7} so therefore, three periods is 3 π / 7. {\displaystyle 3\pi /7.}
Using the roots given in the notes λ 1 = − 2 , λ 2 = 5 {\displaystyle \lambda _{1}=-2,\lambda _{2}=5} , the homogenous solution becomes;y h ( x ) = c 1 e − 2 x + c 2 e 5 x {\displaystyle y_{h}(x)=c_{1}e^{-2x}+c_{2}e^{5x}}
Using initial condtion y ( 0 ) = 1 {\displaystyle y(0)=1} ;1 = c 1 + c 2 {\displaystyle 1=c_{1}+c_{2}} y h ′ ( x ) = − 2 c 1 e − 2 x + 5 c 2 e 5 x {\displaystyle y'_{h}(x)=-2c_{1}e^{-2x}+5c_{2}e^{5x}}
with y ′ ( 0 ) = 0 {\displaystyle y'(0)=0} 0 = − 2 c 1 + 5 c 2 {\displaystyle 0=-2c_{1}+5c_{2}}
Solving for the constants;c 1 = 5 / 7 , c 2 = 2 / 7 {\displaystyle c_{1}=5/7,c_{2}=2/7} y h ( x ) = 5 / 7 e − 2 x + 2 / 7 e 5 x {\displaystyle y_{h}(x)=5/7e^{-2x}+2/7e^{5x}}
Using the y p ( x ) {\displaystyle y_{p}(x)} found in the last part;y = y h + y p {\displaystyle y=y_{h}+y_{p}}
y = 5 / 7 e − 2 x + 2 / 7 e 5 x − 177 / 3922 c o s 7 x − 63 / 3922 s i n 7 x {\displaystyle y=5/7e^{-2x}+2/7e^{5x}-177/3922cos7x-63/3922sin7x}
solved by Luca Imponenti
Problem Statement
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Complete the solution to the following problem
y ″ + 4 y ′ + 13 y = 2 e − 2 x c o s ( 3 x ) {\displaystyle y''+4y'+13y=2e^{-2x}cos(3x)\!} where
y h = e − 2 x [ A c o s ( 3 x ) + B s i n ( 3 x ) ] {\displaystyle y_{h}=e^{-2x}[Acos(3x)+Bsin(3x)]\!}
and
y p = x e − 2 x [ M c o s ( 3 x ) + N s i n ( 3 x ) ] {\displaystyle y_{p}=xe^{-2x}[Mcos(3x)+Nsin(3x)]\!}
Find the overall solution y ( x ) {\displaystyle y(x)\!} corresponds to the initial condition:
y ( 0 ) = 1 , y ′ ( 0 ) = 0 {\displaystyle y(0)=1\ ,\ y'(0)=0\!} Plot the solution over 3 periods.
Particular Solution
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Taking the derivatives of the particular solution y p ( x ) {\displaystyle y_{p}(x)\!}
y p = x e − 2 x [ M c o s ( 3 x ) + N s i n ( 3 x ) ] {\displaystyle y_{p}=xe^{-2x}[Mcos(3x)+Nsin(3x)]\!}
y p ′ = e − 2 x [ s i n ( 3 x ) ( N − 2 N x − 3 M x ) + c o s ( 3 x ) ( 3 N x + M − 2 M x ) ] {\displaystyle y'_{p}=e^{-2x}[sin(3x)(N-2Nx-3Mx)+cos(3x)(3Nx+M-2Mx)]\!}
y p ″ = e − 2 x [ s i n ( 3 x ) ( 12 M x − 6 M − 5 N x − 4 N ) + c o s ( 3 x ) ( 6 N − 5 M x − 4 M − 12 N x ) ] {\displaystyle y''_{p}=e^{-2x}[sin(3x)(12Mx-6M-5Nx-4N)+cos(3x)(6N-5Mx-4M-12Nx)]\!}
Plugging these into the ODE yields
s i n ( 3 x ) ( 6 M − 12 M x + 5 N x + 4 N ) + c o s ( 3 x ) ( 5 M x + 4 M + 12 N x − 6 N ) + {\displaystyle sin(3x)(6M-12Mx+5Nx+4N)+cos(3x)(5Mx+4M+12Nx-6N)+\!}
4 [ s i n ( 3 x ) ( N − 2 N x − 3 M x ) + c o s ( 3 x ) ( 3 N x + M − 2 M x ) ] + {\displaystyle 4[sin(3x)(N-2Nx-3Mx)+cos(3x)(3Nx+M-2Mx)]+}
13 x [ M c o s ( 3 x ) + N s i n ( 3 x ) ] = 2 c o s ( 3 x ) {\displaystyle 13x[Mcos(3x)+Nsin(3x)]=2cos(3x)\!}
Equating like terms allows us to solve for M and N
s i n ( 3 x ) [ ( 12 M x − 6 M − 5 N x − 4 N ) + 4 ( N − 2 N x − 3 M x ) + 13 N x ] = 0 {\displaystyle sin(3x)[(12Mx-6M-5Nx-4N)+4(N-2Nx-3Mx)+13Nx]=0\!}
c o s ( 3 x ) [ ( 6 N − 5 M x − 4 M − 12 N x ) + 4 ( 3 N x + M − 2 M x ) + 13 M x ] = 2 c o s ( 3 x ) {\displaystyle cos(3x)[(6N-5Mx-4M-12Nx)+4(3Nx+M-2Mx)+13Mx]=2cos(3x)\!}
− 6 M = 0 {\displaystyle -6M=0\!}
6 N = 2 {\displaystyle 6N=2\!}
M = 0 , N = 1 3 {\displaystyle M=0\ ,\ N={\frac {1}{3}}\!}
So the particular solution is
y p = 1 3 x e − 2 x s i n ( 3 x ) {\displaystyle y_{p}={\frac {1}{3}}xe^{-2x}sin(3x)\!}
Overall Solution
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The overall solution in the sum of the homogeneous and particular solutions
y ( x ) = y h ( x ) + y p ( x ) {\displaystyle y(x)=y_{h}(x)+y_{p}(x)\!}
y ( x ) = e − 2 x [ A c o s ( 3 x ) + B s i n ( 3 x ) ] + 1 3 x e − 2 x s i n ( 3 x ) {\displaystyle y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]+{\frac {1}{3}}xe^{-2x}sin(3x)\!}
To find A and B we apply the initial conditions
y ( 0 ) = 1 , y ′ ( 0 ) = 0 {\displaystyle y(0)=1\ ,\ y'(0)=0\!}
y ( 0 ) = 1 = A {\displaystyle y(0)=1=A\!}
Taking the derivative
y ′ ( x ) = d d x [ e − 2 x [ c o s ( 3 x ) + B s i n ( 3 x ) ] + 1 3 x e − 2 x s i n ( 3 x ) ] {\displaystyle y'(x)={\frac {d}{dx}}[e^{-2x}[cos(3x)+Bsin(3x)]+{\frac {1}{3}}xe^{-2x}sin(3x)]\!}
y ′ ( x ) = e − 2 x [ ( 3 B + x − 2 ) c o s ( 3 x ) − ( 2 B + 2 3 x + 8 3 ) s i n ( 3 x ) ] {\displaystyle y'(x)=e^{-2x}[(3B+x-2)cos(3x)-(2B+{\frac {2}{3}}x+{\frac {8}{3}})sin(3x)]\!}
y ′ ( 0 ) = 0 = 3 B − 2 {\displaystyle y'(0)=0=3B-2\!}
B = 2 3 {\displaystyle B={\frac {2}{3}}\!}
Giving us the overall solution
y ( x ) = e − 2 x [ c o s ( 3 x ) + 2 3 s i n ( 3 x ) + 1 3 x s i n ( 3 x ) ] {\displaystyle y(x)=e^{-2x}[cos(3x)+{\frac {2}{3}}sin(3x)+{\frac {1}{3}}xsin(3x)]\!}
The period for c o s ( 3 x ) , s i n ( 3 x ) {\displaystyle cos(3x)\ ,\ sin(3x)\!} is 2 π 3 {\displaystyle {\frac {2\pi }{3}}}
Plotting the solution y ( x ) {\displaystyle y(x)\!} over 3 periods yields
Solved by Daniel Suh
Problem Statement
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v = 4 e 1 + 2 e 2 = c 1 b 1 + c 2 b 2 {\displaystyle v=4e_{1}+2e_{2}=c_{1}b_{1}+c{2}b_{2}\!}
b 1 = 2 e 1 + 7 e 2 {\displaystyle b_{1}=2e_{1}+7e_{2}\!}
b 2 = 1.5 e 1 + 3 e 2 {\displaystyle b_{2}=1.5e_{1}+3e_{2}\!}
1. Find the components c 1 , c 2 {\displaystyle c_{1},c_{2}\!} using the Gram matrix.
2. Verify the result by using b 1 {\displaystyle b_{1}\!} and b 2 {\displaystyle b_{2}\!} , and rely on the non-zero determinant matrix of b 1 {\displaystyle b_{1}\!} and b 2 {\displaystyle b_{2}\!} relative to the bases of e 1 {\displaystyle e_{1}\!} and e 2 {\displaystyle e_{2}\!} .
Part 1 Solution
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Gram Matrix
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T ( b 1 , b 2 ) = [ < b 1 , b 1 > < b 1 , b 2 > < b 2 , b 1 > < b 2 , b 2 > ] {\displaystyle T(b_{1},b_{2})={\begin{bmatrix}<b_{1},b_{1}>&<b_{1},b_{2}>\\<b_{2},b_{1}>&<b_{2},b_{2}>\end{bmatrix}}} < b i , b j >=< b i ⋅ b j > {\displaystyle <b_{i},b_{j}>=<b_{i}\cdot b_{j}>\!}
Thus,
< b 1 , b 1 >=< b 1 ⋅ b 1 >=< ( 2 ) ( 2 ) + ( 7 ) ( 7 ) >= 53 {\displaystyle <b_{1},b_{1}>=<b_{1}\cdot b_{1}>=<(2)(2)+(7)(7)>=53\!}
< b 2 , b 2 >=< b 2 ⋅ b 2 >=< ( 1.5 ) ( 1.5 ) + ( 3 ) ( 3 ) >= 11.25 {\displaystyle <b_{2},b_{2}>=<b_{2}\cdot b_{2}>=<(1.5)(1.5)+(3)(3)>=11.25\!}
< b 1 , b 2 >=< b 2 , b 1 >=< b 1 ⋅ b 2 >=< ( 2 ) ( 1.5 ) + ( 7 ) ( 3 ) >= 24 {\displaystyle <b_{1},b_{2}>=<b_{2},b_{1}>=<b_{1}\cdot b_{2}>=<(2)(1.5)+(7)(3)>=24\!}
T ( b 1 , b 2 ) = [ 53 24 24 11.25 ] {\displaystyle T(b_{1},b_{2})={\begin{bmatrix}53&24\\24&11.25\end{bmatrix}}}
Γ = d e t [ T ] = ( 53 ) ( 11.25 ) − ( 24 ) ( 24 ) = 20.25 {\displaystyle \Gamma =det[T]=(53)(11.25)-(24)(24)=20.25\!}
Defining c
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Define:
c = [ c 1 c 2 ] {\displaystyle c={\begin{bmatrix}c_{1}\\c_{2}\end{bmatrix}}} d = [ < b 1 , v > < b 2 , v > ] = [ d 1 d 2 ] {\displaystyle d={\begin{bmatrix}<b_{1},v>\\<b_{2},v>\end{bmatrix}}={\begin{bmatrix}d_{1}\\d_{2}\end{bmatrix}}\!}
If Γ ≠ 0 {\displaystyle \Gamma \neq 0\!} , then Γ − 1 {\displaystyle \Gamma ^{-1}\!} exists
c = Γ − 1 d {\displaystyle c=\Gamma ^{-1}d\!}
Finding c
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Γ = 20.25 ≠ 0 {\displaystyle \Gamma =20.25\neq 0\!} thus, Γ − 1 {\displaystyle \Gamma ^{-1}\!} exists
d = [ d 1 d 2 ] = [ ( 2 ) ( 4 ) + ( 7 ) ( 2 ) ( 1.5 ) ( 4 ) + ( 3 ) ( 2 ) ] = [ 22 12 ] {\displaystyle d={\begin{bmatrix}d_{1}\\d_{2}\end{bmatrix}}={\begin{bmatrix}(2)(4)+(7)(2)\\(1.5)(4)+(3)(2)\end{bmatrix}}={\begin{bmatrix}22\\12\end{bmatrix}}\!}
[ c 1 c 2 ] = [ 53 24 24 11.25 ] − 1 [ 22 12 ] {\displaystyle {\begin{bmatrix}c_{1}\\c_{2}\end{bmatrix}}={\begin{bmatrix}53&24\\24&11.25\end{bmatrix}}^{-1}{\begin{bmatrix}22\\12\end{bmatrix}}\!}
[ c 1 c 2 ] = [ − 2 5.33 ] {\displaystyle {\begin{bmatrix}c_{1}\\c_{2}\end{bmatrix}}={\begin{bmatrix}-2\\5.33\end{bmatrix}}\!}
Part 2 Solution
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v = 4 e 1 + 2 e 2 ≡ c 1 b 1 + c 2 b 2 {\displaystyle v=4e_{1}+2e_{2}\equiv c_{1}b_{1}+c_{2}b_{2}\!}
c 1 b 1 + c 2 b 2 = ( − 2 ) ( 2 e 1 + 7 e 2 ) + ( 5.33 ) ( 1.5 e 1 + 3 e 2 ) {\displaystyle c_{1}b_{1}+c_{2}b_{2}=(-2)(2e_{1}+7e_{2})+(5.33)(1.5e_{1}+3e_{2})\!}
c 1 b 1 + c 2 b 2 = − 4 e 1 − 14 e 2 + 8 e 1 + 16 e 2 {\displaystyle c_{1}b_{1}+c_{2}b_{2}=-4e_{1}-14e_{2}+8e_{1}+16e_{2}\!}
c 1 b 1 + c 2 b 2 = 4 e 1 + 2 e 2 {\displaystyle c_{1}b_{1}+c_{2}b_{2}=4e_{1}+2e_{2}\!}
v = 4 e 1 + 2 e 2 ≡ c 1 b 1 + c 2 b 2 {\displaystyle v=4e_{1}+2e_{2}\equiv c_{1}b_{1}+c_{2}b_{2}\!}
∴ {\displaystyle \therefore \!} solution is correct
Problem Statement
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Find the integral ∫ x n l o g ( 1 + x ) d x {\displaystyle \int x^{n}log(1+x)dx\!} for n = 0 {\displaystyle n=0\!} and n = 1 {\displaystyle n=1\!}
Using integration by parts, and then with the help of of
General Binomial Theorem( x + y ) n = ∑ k = 0 n ( n k ) x n − k y k {\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{n-k}y^{k}\!}
Solution
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For n = 0 {\displaystyle n=0\!} :∫ x 0 l o g ( 1 + x ) d x = ∫ l o g ( 1 + x ) d x {\displaystyle \int x^{0}log(1+x)dx=\int log(1+x)dx\!}
For substitution by parts, u = l o g ( 1 + x ) , d u = 1 1 + x , d v = d x , v = x {\displaystyle u=log(1+x),du={\frac {1}{1+x}},dv=dx,v=x\!} ∫ l o g ( 1 + x ) d x = x l o g ( 1 + x ) − ∫ x 1 + x d x {\displaystyle \int log(1+x)dx=xlog(1+x)-\int {\frac {x}{1+x}}dx\!} ∫ l o g ( 1 + x ) d x = x l o g ( 1 + x ) − ∫ ( 1 − 1 1 + x ) d x {\displaystyle \int log(1+x)dx=xlog(1+x)-\int (1-{\frac {1}{1+x}})dx\!} ∫ l o g ( 1 + x ) d x = x l o g ( 1 + x ) − x + l o g ( 1 + x ) + C {\displaystyle \int log(1+x)dx=xlog(1+x)-x+log(1+x)+C\!}
Therefore:
∫ l o g ( 1 + x ) d x = ( x + 1 ) l o g ( 1 + x ) − x + C {\displaystyle \int log(1+x)dx=(x+1)log(1+x)-x+C\!}
Using the General Binomial Theorem:( x + y ) 0 = ∑ k = 0 0 ( 0 k ) x 0 − k y k = 1 {\displaystyle (x+y)^{0}=\sum _{k=0}^{0}{\binom {0}{k}}x^{0-k}y^{k}=1\!}
Therefore:
∫ ( 1 ) l o g ( 1 + x ) d x = ∫ l o g ( 1 + x ) d x {\displaystyle \int (1)log(1+x)dx=\int log(1+x)dx\!}
Which we have previously found that answer as:
∫ l o g ( 1 + x ) d x = ( x + 1 ) l o g ( 1 + x ) − x + C {\displaystyle \int log(1+x)dx=(x+1)log(1+x)-x+C\!}
For n = 1 {\displaystyle n=1\!} :∫ x 1 l o g ( 1 + x ) d x = ∫ x l o g ( 1 + x ) d x {\displaystyle \int x^{1}log(1+x)dx=\int xlog(1+x)dx\!}
Initially we use the following substitutions: t = 1 + x , x = t − 1 , d t = d x {\displaystyle t=1+x,x=t-1,dt=dx\!} ∫ x l o g ( 1 + x ) d x = ∫ ( t − 1 ) l o g ( t ) d t = ∫ ( t l o g ( t ) − log ( t ) ) d t {\displaystyle \int xlog(1+x)dx=\int (t-1)log(t)dt=\int (tlog(t)-\log(t))dt\!}
First let us consider the first term: ∫ t l o g ( t ) d t {\displaystyle \int tlog(t)dt\!}
Next, we use the integration by parts: u = log t , d u = 1 t d t , d v = t d t , v = 1 2 t 2 {\displaystyle u=\log {t},du={\frac {1}{t}}dt,dv=tdt,v={\frac {1}{2}}t^{2}\!} ∫ t l o g ( t ) d t = 1 2 t 2 l o g ( t ) − ∫ 1 2 t 2 ( 1 t d t ) {\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-\int {\frac {1}{2}}t^{2}({\frac {1}{t}}dt)\!} ∫ t l o g ( t ) d t = 1 2 t 2 l o g ( t ) − ∫ 1 2 t d t ) {\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-\int {\frac {1}{2}}tdt)\!} ∫ t l o g ( t ) d t = 1 2 t 2 l o g ( t ) − 1 4 t 2 {\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}\!}
Next let us consider the second term: ∫ l o g ( t ) d t {\displaystyle \int log(t)dt\!}
Again, we will use integration by parts: u = log t , d u = 1 t d t , d v = d t , v = t {\displaystyle u=\log {t},du={\frac {1}{t}}dt,dv=dt,v=t\!} ∫ t l o g ( t ) d t = t l o g ( t ) − ∫ t ( 1 t d t ) {\displaystyle \int tlog(t)dt=tlog(t)-\int t({\frac {1}{t}}dt)\!} ∫ t l o g ( t ) d t = t l o g ( t ) − ∫ d t {\displaystyle \int tlog(t)dt=tlog(t)-\int dt\!} ∫ t l o g ( t ) d t = t l o g ( t ) − t {\displaystyle \int tlog(t)dt=tlog(t)-t\!}
Therefore:∫ ( t l o g ( t ) − log ( t ) ) d t = 1 2 t 2 l o g ( t ) − 1 4 t 2 − ( t l o g ( t ) − t ) {\displaystyle \int (tlog(t)-\log(t))dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}-(tlog(t)-t)\!} ∫ ( t l o g ( t ) − log ( t ) ) d t = 1 2 t 2 l o g ( t ) − 1 4 t 2 − t l o g ( t ) + t {\displaystyle \int (tlog(t)-\log(t))dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}-tlog(t)+t\!}
Re-substituting for t: ∫ x l o g ( 1 + x ) d x = 1 2 ( 1 + x ) 2 l o g ( 1 + x ) − 1 4 ( 1 + x ) 2 − ( 1 + x ) l o g ( 1 + x ) + ( 1 + x ) + C {\displaystyle \int xlog(1+x)dx={\frac {1}{2}}(1+x)^{2}log(1+x)-{\frac {1}{4}}(1+x)^{2}-(1+x)log(1+x)+(1+x)+C\!} ∫ x l o g ( 1 + x ) d x = ( 1 + x ) ( 1 2 ( 1 + x ) l o g ( 1 + x ) − 1 4 ( 1 + x ) − l o g ( 1 + x ) + 1 ) + C {\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}(1+x)log(1+x)-{\frac {1}{4}}(1+x)-log(1+x)+1)+C\!} ∫ x l o g ( 1 + x ) d x = ( 1 + x ) ( 1 2 x l o g ( 1 + x ) − 1 2 l o g ( 1 + x ) − 1 4 x + 3 4 ) + C {\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!}
Therefore:
∫ x l o g ( 1 + x ) d x = ( 1 + x ) ( 1 2 x l o g ( 1 + x ) − 1 2 l o g ( 1 + x ) − 1 4 x + 3 4 ) + C {\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!}
Using the General Binomial Theorem for the integral with t substitution ∫ ( t − 1 ) l o g ( t ) d t {\displaystyle \int (t-1)log(t)dt\!} : ( x + y ) 1 = ( t + ( − 1 ) ) 1 = ( t + ( − 1 ) ) = ∑ k = 0 1 ( 1 k ) x 1 − k y k = ∑ k = 0 1 ( 1 k ) t 1 − k ( − 1 ) k = t − 1 {\displaystyle (x+y)^{1}=(t+(-1))^{1}=(t+(-1))=\sum _{k=0}^{1}{\binom {1}{k}}x^{1-k}y^{k}=\sum _{k=0}^{1}{\binom {1}{k}}t^{1-k}(-1)^{k}=t-1\!}
Therefore:
∫ ( t − 1 ) l o g ( t ) d t = ∫ x l o g ( 1 + x ) d x {\displaystyle \int (t-1)log(t)dt=\int xlog(1+x)dx\!}
Which we have previously found that answer as:
∫ x l o g ( 1 + x ) d x = ( 1 + x ) ( 1 2 x l o g ( 1 + x ) − 1 2 l o g ( 1 + x ) − 1 4 x + 3 4 ) + C {\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!}
Solved by: Gonzalo Perez
Problem Statement
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Consider the L2-ODE-CC (5) p.7b-7 with l o g ( 1 + x ) {\displaystyle log(1+x)\!} as excitation:
y ″ − 3 y ′ + 2 y = r ( x ) {\displaystyle y''-3y'+2y=r(x)\!} (5) p.7b-7
r ( x ) = l o g ( 1 + x ) {\displaystyle r(x)=log(1+x)\!} (1) p.7c-28
and the initial conditions
y ( − 3 4 ) = 1 , y ′ ( − 3 4 ) = 0 {\displaystyle y({\frac {-3}{4}})=1,y'({\frac {-3}{4}})=0\!} .
Project the excitation r ( x ) {\displaystyle r(x)\!} on the polynomial basis
b j ( x ) = x j , j = 0 , 1 , . . . , n {\displaystyle {b_{j}(x)=x^{j},j=0,1,...,n}\!} (1)
i.e., find d j {\displaystyle d_{j}\!} such that
r ( x ) ≈ r n ( x ) = ∑ j = 0 n d j x j {\displaystyle r(x)\approx r_{n}(x)=\sum _{j=0}^{n}d_{j}x^{j}\!} (2)
for x in [ − 3 4 , 3 ] {\displaystyle [{\frac {-3}{4}},3]\!} , and for n = 3, 6, 9.
Plot r ( x ) {\displaystyle r(x)\!} and r n ( x ) {\displaystyle r_{n}(x)\!} to show uniform approximation and convergence.
Note that:
⟨ x i , r ⟩ = ∫ a b x i l o g ( 1 + x ) d x {\displaystyle \left\langle x^{i},r\right\rangle =\int _{a}^{b}x^{i}log(1+x)dx\!} (3)
Solution
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To solve this problem, it is important to know that the scalar product is defined as the following:
⟨ b 0 , b 0 ⟩ = ∫ x 0 ⋅ x 0 d x {\displaystyle \left\langle b_{0},b_{0}\right\rangle =\int x^{0}\cdot x^{0}dx\!} .
Therefore, it follows that:
⟨ b i , b j ⟩ = ∫ x i ⋅ x j d x {\displaystyle \left\langle b_{i},b_{j}\right\rangle =\int x^{i}\cdot x^{j}dx\!} , where b i ( x ) = x i {\displaystyle b_{i}(x)=x^{i}\!} and b j ( x ) = x j {\displaystyle b_{j}(x)=x^{j}\!} .
We know that if b 1 , b 2 {\displaystyle b_{1},b_{2}\!} are linearly independent, then by theorem on p.7c-37, the matrix is solvable.
According to this and (3)p.8-14:
If Γ ≠ 0 ⇒ Γ − 1 {\displaystyle \Gamma \neq 0\Rightarrow \Gamma ^{-1}\!} exists ⇒ c = Γ − 1 d {\displaystyle \Rightarrow c=\Gamma ^{-1}d\!} . (3)p.8-14
Now let's define the Gram matrix Γ {\displaystyle \Gamma \!} as a function of b i {\displaystyle b_{i}\!} :
Γ ( b i ) = [ ⟨ b 0 , b 0 ⟩ ⟨ b 0 , b 1 ⟩ . . . ⟨ b 0 , b n ⟩ . . . . . . . . . . . . . . . . . . . . . . . . ⟨ b n , b 0 ⟩ ⟨ b n , b 1 ⟩ . . . ⟨ b n , b n ⟩ ] {\displaystyle \Gamma (b_{i})={\begin{bmatrix}\left\langle b_{0},b_{0}\right\rangle &\left\langle b_{0},b_{1}\right\rangle &...&\left\langle b_{0},b_{n}\right\rangle \\...&...&...&...\\...&...&...&...\\\left\langle b_{n},b_{0}\right\rangle &\left\langle b_{n},b_{1}\right\rangle &...&\left\langle b_{n},b_{n}\right\rangle \end{bmatrix}}\!} (1)p.8-13
Defining the "d" matrix as was done in (3)p.8-13, we get:
d = { ⟨ b 0 , r ⟩ ⟨ b 1 , r ⟩ . . . ⟨ b n , r ⟩ } {\displaystyle d={\begin{Bmatrix}\left\langle b_{0},r\right\rangle \\\left\langle b_{1},r\right\rangle \\...\\\left\langle b_{n},r\right\rangle \end{Bmatrix}}\!} . (3)p.8-13
And according to (1)p.8-15: r n ( x ) = ∑ 0 n c i x i {\displaystyle r_{n}(x)=\sum _{0}^{n}c_{i}x^{i}\!} (1)p.8-15
Now, we can find the values to compare r n {\displaystyle r_{n}\!} to y {\displaystyle y\!} .
Using Matlab, this is the code that was used to produce the results:
The Matlab code above produced the following graph:
Where r n ( x ) {\displaystyle r_{n}(x)\!} is represented by the dashed line and the approximation,y ( x ) {\displaystyle y(x)\!} , is represented by the red line. This code can work for all n values.
In a seperate series of plots, compare the approximation of the function log ( x + 1 ) {\displaystyle \log(x+1)\!} by Taylor series expansion about x = 0 {\displaystyle x=0\!} .
Where: f ( x ) = ∑ n = 0 ∞ f ( n ) ( x ^ ) n ! ( x − x ^ ) n {\displaystyle f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}({\hat {x}})}{n!}}(x-{\hat {x}})^{n}\!}
Solution
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For n=1:
log ( x + 1 ) = x log ( 10 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}\!}
For n=2:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}\!}
For n=3:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}\!}
For n=4:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}\!}
For n=5:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}\!}
For n=6:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}\!}
For n=7:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}\!}
For n=8:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}\!}
For n=9:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) + x 9 log ( 10 9 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}+{\frac {x^{9}}{\log(10^{9})}}\!}
For n=10:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) + x 9 log ( 10 9 ) − x 10 log ( 10 10 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}+{\frac {x^{9}}{\log(10^{9})}}-{\frac {x^{10}}{\log(10^{10})}}\!}
For n=11:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) + x 9 log ( 10 9 ) − x 10 log ( 10 10 ) + x 11 log ( 10 11 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}+{\frac {x^{9}}{\log(10^{9})}}-{\frac {x^{10}}{\log(10^{10})}}+{\frac {x^{11}}{\log(10^{11})}}\!}
For n=12:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) + x 9 log ( 10 9 ) − x 10 log ( 10 10 ) + x 11 log ( 10 11 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}+{\frac {x^{9}}{\log(10^{9})}}-{\frac {x^{10}}{\log(10^{10})}}+{\frac {x^{11}}{\log(10^{11})}}\!}
− x 12 log ( 10 12 ) {\displaystyle -{\frac {x^{12}}{\log(10^{12})}}\!}
For n=13:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) + x 9 log ( 10 9 ) − x 10 log ( 10 10 ) + x 11 log ( 10 11 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}+{\frac {x^{9}}{\log(10^{9})}}-{\frac {x^{10}}{\log(10^{10})}}+{\frac {x^{11}}{\log(10^{11})}}\!}
− x 12 log ( 10 12 ) + x 13 log ( 10 13 ) {\displaystyle -{\frac {x^{12}}{\log(10^{12})}}+{\frac {x^{13}}{\log(10^{13})}}\!}
For n=14:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) + x 9 log ( 10 9 ) − x 10 log ( 10 10 ) + x 11 log ( 10 11 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}+{\frac {x^{9}}{\log(10^{9})}}-{\frac {x^{10}}{\log(10^{10})}}+{\frac {x^{11}}{\log(10^{11})}}\!}
− x 12 log ( 10 12 ) + x 13 log ( 10 13 ) − x 14 log ( 10 14 ) {\displaystyle -{\frac {x^{12}}{\log(10^{12})}}+{\frac {x^{13}}{\log(10^{13})}}-{\frac {x^{14}}{\log(10^{14})}}\!}
For n=15:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) + x 9 log ( 10 9 ) − x 10 log ( 10 10 ) + x 11 log ( 10 11 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}+{\frac {x^{9}}{\log(10^{9})}}-{\frac {x^{10}}{\log(10^{10})}}+{\frac {x^{11}}{\log(10^{11})}}\!}
− x 12 log ( 10 12 ) + x 13 log ( 10 13 ) − x 14 log ( 10 14 ) + x 15 log ( 10 15 ) {\displaystyle -{\frac {x^{12}}{\log(10^{12})}}+{\frac {x^{13}}{\log(10^{13})}}-{\frac {x^{14}}{\log(10^{14})}}+{\frac {x^{15}}{\log(10^{15})}}\!}
For n=16:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) + x 9 log ( 10 9 ) − x 10 log ( 10 10 ) + x 11 log ( 10 11 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}+{\frac {x^{9}}{\log(10^{9})}}-{\frac {x^{10}}{\log(10^{10})}}+{\frac {x^{11}}{\log(10^{11})}}\!}
− x 12 log ( 10 12 ) + x 13 log ( 10 13 ) − x 14 log ( 10 14 ) + x 15 log ( 10 15 ) − x 16 log ( 10 16 ) {\displaystyle -{\frac {x^{12}}{\log(10^{12})}}+{\frac {x^{13}}{\log(10^{13})}}-{\frac {x^{14}}{\log(10^{14})}}+{\frac {x^{15}}{\log(10^{15})}}-{\frac {x^{16}}{\log(10^{16})}}\!}
Using Matlab to plot the graph:
Find y n ( x ) {\displaystyle y_{n}(x)\!} such that:
y n ″ + a y n ′ + b y n = r n ( x ) {\displaystyle y_{n}''+ay_{n}'+by_{n}=r_{n}(x)\!} (1) p.7c-27
with the same initial conditions as in (2) p.7c-28.
Plot y n ( x ) {\displaystyle y_{n}(x)\!} for n = 3, 6, 9, for x in [ − 3 4 , 3 ] {\displaystyle [{\frac {-3}{4}},3]\!} .
In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.
Solution
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First, we find the homogeneous solution to the ODE:
The characteristic equation is:λ 2 − 3 λ + 2 = 0 {\displaystyle \lambda ^{2}-3\lambda +2=0\!} ( λ − 2 ) ( λ − 1 ) = 0 {\displaystyle (\lambda -2)(\lambda -1)=0\!}
Then,
λ = 1 , 2 {\displaystyle \lambda =1,2\!}
Therefore the homogeneous solution is:y h = C 1 e ( 2 x ) + c 2 e x {\displaystyle y_{h}=C_{1}e^{(}2x)+c_{2}e^{x}\!}
Now to find the particulate solution
For n=3:
r ( x ) = ∑ 0 n − ( − 1 ) n x n n ln ( 10 ) {\displaystyle r(x)=\sum _{0}^{n}-{\frac {(-1)^{n}x^{n}}{n\ln(10)}}\!}
r ( x ) = x ln ( 10 ) − x 2 2 ln ( 10 ) + x 3 3 ln ( 10 ) {\displaystyle r(x)={\frac {x}{\ln(10)}}-{\frac {x^{2}}{2\ln(10)}}+{\frac {x^{3}}{3\ln(10)}}\!}
We can then use a matrix to organize the known coefficients:
[ 2 − 3 2 0 0 0 2 − 6 6 0 0 0 2 − 9 12 0 0 0 2 − 12 0 0 0 0 2 ] [ K 0 K 1 K 2 K 3 ] = [ 0 1 l n ( 10 ) 1 2 l n ( 10 ) 1 3 l n ( 10 ) ] {\displaystyle {\begin{bmatrix}2&-3&2&0&0\\0&2&-6&6&0\\0&0&2&-9&12\\0&0&0&2&-12\\0&0&0&0&2\\\end{bmatrix}}{\begin{bmatrix}K_{0}\\K_{1}\\K_{2}\\K_{3}\end{bmatrix}}={\begin{bmatrix}0\\{\frac {1}{ln(10)}}\\{\frac {1}{2ln(10)}}\\{\frac {1}{3ln(10)}}\end{bmatrix}}\!}
Then, using MATLAB and the backlash operator we can solve for these unknowns:
Thereforey p 4 = 4.0444 + 3.7458 x + 1.5743 x 2 + 0.3981 x 3 + 0.0543 x 4 {\displaystyle y_{p4}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}\!}
Superposing the homogeneous and particulate solution we gety n = 4.0444 + 3.7458 x + 1.5743 x 2 + 0.3981 x 3 + 0.0543 x 4 + C 1 e 2 x + C 2 e x {\displaystyle y_{n}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}+C_{1}e^{2x}+C_{2}e^{x}\!}
Differentiating:y n ′ = 3.7458 + 3.1486 x + 1.1943 x 2 + 0.2172 x 3 + 2 C 1 e 2 x + C 2 e x {\displaystyle y'_{n}=3.7458+3.1486x+1.1943x^{2}+0.2172x^{3}+2C_{1}e^{2x}+C_{2}e^{x}\!}
Evaluating at the initial conditions:y ( − 0.75 ) = 0.9698261719 + 0.231301601 C ! + 0.4723665527 C 2 = 1 {\displaystyle y(-0.75)=0.9698261719+0.231301601C_{!}+0.4723665527C_{2}=1\!} y ′ ( − 0.75 ) = 1.9645125 + 0.4462603203 C 1 + 0.4723665527 C 2 = 0 {\displaystyle y'(-0.75)=1.9645125+0.4462603203C_{1}+0.4723665527C_{2}=0\!}
We obtain:C 1 = − 4.46 {\displaystyle C_{1}=-4.46\!} C 2 = 0.055 {\displaystyle C_{2}=0.055\!}
Finally we have:y n = 4.0444 + 3.7458 x + 1.5743 x 2 + 0.3981 x 3 + 0.0543 x 4 − 4.46 e 2 x + 0.055 e x {\displaystyle y_{n}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}-4.46e^{2x}+0.055e^{x}\!}
For n=6:
r ( x ) = ∑ 0 n − ( − 1 ) n x n n ln ( 10 ) {\displaystyle r(x)=\sum _{0}^{n}-{\frac {(-1)^{n}x^{n}}{n\ln(10)}}\!}
r ( x ) = x ln ( 10 ) − x 2 2 ln ( 10 ) + x 3 3 ln ( 10 ) − x 4 4 ln ( 10 ) + x 5 5 ln ( 10 ) − x 6 6 ln ( 10 ) {\displaystyle r(x)={\frac {x}{\ln(10)}}-{\frac {x^{2}}{2\ln(10)}}+{\frac {x^{3}}{3\ln(10)}}-{\frac {x^{4}}{4\ln(10)}}+{\frac {x^{5}}{5\ln(10)}}-{\frac {x^{6}}{6\ln(10)}}\!}
We can then use a matrix to organize the known coefficients:
[ 2 − 3 2 0 0 0 0 0 0 2 − 6 6 0 0 0 0 0 0 2 − 9 12 0 0 0 0 0 0 2 − 12 20 0 0 0 0 0 0 2 − 15 30 0 0 0 0 0 0 2 − 18 42 0 0 0 0 0 0 2 − 21 0 0 0 0 0 0 0 2 ] [ K 0 K 1 K 2 K 3 K 4 K 5 K 6 ] [ 0 1 l n ( 10 ) 1 2 l n ( 10 ) 1 3 l n ( 10 ) 1 4 l n ( 10 ) 1 5 l n ( 10 ) 1 6 l n ( 10 ) ] {\displaystyle {\begin{bmatrix}2&-3&2&0&0&0&0&0\\0&2&-6&6&0&0&0&0\\0&0&2&-9&12&0&0&0\\0&0&0&2&-12&20&0&0\\0&0&0&0&2&-15&30&0\\0&0&0&0&0&2&-18&42\\0&0&0&0&0&0&2&-21\\0&0&0&0&0&0&0&2\\\end{bmatrix}}{\begin{bmatrix}K_{0}\\K_{1}\\K_{2}\\K_{3}\\K_{4}\\K_{5}\\K_{6}\end{bmatrix}}{\begin{bmatrix}0\\{\frac {1}{ln(10)}}\\{\frac {1}{2ln(10)}}\\{\frac {1}{3ln(10)}}\\{\frac {1}{4ln(10)}}\\{\frac {1}{5ln(10)}}\\{\frac {1}{6ln(10)}}\end{bmatrix}}\!}
Then, using MATLAB and the backlash operator we can solve for these unknowns:
Thereforey p 7 = 377.4833 + 375.3933 x + 185.6066 x 2 + 60.5479 x 3 + 14.4946 x 4 + 2.6492 x 5 + 0.3619 x 6 + 0.0310 x 7 {\displaystyle y_{p7}=377.4833+375.3933x+185.6066x^{2}+60.5479x^{3}+14.4946x^{4}+2.6492x^{5}+0.3619x^{6}+0.0310x^{7}\!}
Superposing the homogeneous and particulate solution we gety n = 377.4833 + 375.3933 x + 185.6066 x 2 + 60.5479 x 3 + 14.4946 x 4 + 2.6492 x 5 + 0.3619 x 6 + 0.0310 x 7 + C 1 e 2 x + c 2 e x {\displaystyle y_{n}=377.4833+375.3933x+185.6066x^{2}+60.5479x^{3}+14.4946x^{4}+2.6492x^{5}+0.3619x^{6}+0.0310x^{7}+C_{1}e^{2x}+c_{2}e^{x}\!}
Differentiating:y n ′ = 375.3933 + 371.213 x + 181.644 x 2 + 57.9784 x 3 + 13.46 x 4 + 2.1714 x 5 + 0.214 x 6 + 2 C 1 e 2 x + C 2 e x {\displaystyle y'_{n}=375.3933+371.213x+181.644x^{2}+57.9784x^{3}+13.46x^{4}+2.1714x^{5}+0.214x^{6}+2C_{1}e^{2x}+C_{2}e^{x}\!}
Evaluating at the initial conditions:y ( − 0.75 ) = 178.816 + 0.2231301601 C ! + 0.4723665527 C 2 = 1 {\displaystyle y(-0.75)=178.816+0.2231301601C_{!}+0.4723665527C_{2}=1\!} y ′ ( − 0.75 ) = 178.413 + 0.4462603203 C 1 + 0.4723665527 C 2 {\displaystyle y'(-0.75)=178.413+0.4462603203C_{1}+0.4723665527C_{2}\!}
We obtain:C 1 = − 2.6757 {\displaystyle C_{1}=-2.6757\!} C 2 = − 375.173 {\displaystyle C_{2}=-375.173\!}
Finallyy n = 377.4833 + 375.3933 x + 185.6066 x 2 + 60.5479 x 3 + 14.4946 x 4 + 2.6492 x 5 + 0.3619 x 6 + 0.0310 x 7 + − 2.6757 e 2 x − 375.173 e x {\displaystyle y_{n}=377.4833+375.3933x+185.6066x^{2}+60.5479x^{3}+14.4946x^{4}+2.6492x^{5}+0.3619x^{6}+0.0310x^{7}+-2.6757e^{2x}-375.173e^{x}\!}
For n=9:
r ( x ) = ∑ 0 n − ( − 1 ) n x n n ln ( 10 ) {\displaystyle r(x)=\sum _{0}^{n}-{\frac {(-1)^{n}x^{n}}{n\ln(10)}}\!}
r ( x ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) + x 9 log ( 10 9 ) {\displaystyle r(x)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}+{\frac {x^{9}}{\log(10^{9})}}\!}
We can then use a matrix to organize the known coefficients:
[ K 0 K 1 K 2 K 3 K 4 K 5 K 6 K 7 K 8 K 9 ] = [ 0 1 l n ( 10 ) 1 2 l n ( 10 ) 1 3 l n ( 10 ) 1 4 l n ( 10 ) 1 5 l n ( 10 ) 1 6 l n ( 10 ) 1 7 l n ( 10 ) 1 8 l n ( 10 ) 1 9 l n ( 10 ) ] {\displaystyle {\begin{bmatrix}K_{0}\\K_{1}\\K_{2}\\K_{3}\\K_{4}\\K_{5}\\K_{6}\\K_{7}\\K_{8}\\K_{9}\end{bmatrix}}={\begin{bmatrix}0\\{\frac {1}{ln(10)}}\\{\frac {1}{2ln(10)}}\\{\frac {1}{3ln(10)}}\\{\frac {1}{4ln(10)}}\\{\frac {1}{5ln(10)}}\\{\frac {1}{6ln(10)}}\\{\frac {1}{7ln(10)}}\\{\frac {1}{8ln(10)}}\\{\frac {1}{9ln(10)}}\end{bmatrix}}\!}
Then, using MATLAB and the backlash operator we can solve for these unknowns:
Thereforey p 11 = 1753158.594 + 1752673.419 x + 875851.535 x 2 + 291627.134 x 3 + 72745.1129 x 4 + 14484.362 x 5 + 2392.510 x 6 + 335.632 x 7 + 40.417 x 8 {\displaystyle y_{p11}=1753158.594+1752673.419x+875851.535x^{2}+291627.134x^{3}+72745.1129x^{4}+14484.362x^{5}+2392.510x^{6}+335.632x^{7}+40.417x^{8}\!} + 4.1499 x 9 + 0.3474 x ( 10 ) + 0.0197 x ( 11 ) {\displaystyle +4.1499x^{9}+0.3474x^{(}10)+0.0197x^{(}11)\!}
Superposing the homogeneous and particulate solution we gety n = 1753158.594 + 1752673.419 x + 875851.535 x 2 + 291627.134 x 3 + 72745.1129 x 4 + 14484.362 x 5 + 2392.510 x 6 + 335.632 x 7 + 40.417 x 8 {\displaystyle y_{n}=1753158.594+1752673.419x+875851.535x^{2}+291627.134x^{3}+72745.1129x^{4}+14484.362x^{5}+2392.510x^{6}+335.632x^{7}+40.417x^{8}\!} + 4.1499 x 9 + 0.3474 x ( 10 ) + 0.0197 x ( 11 ) + C 1 e 2 x + C 2 e ( x ) {\displaystyle +4.1499x^{9}+0.3474x^{(}10)+0.0197x^{(}11)+C_{1}e^{2x}+C_{2}e^{(}x)\!}
Differentiating:y n ′ = 0.2167 x 1 0 + 3.474 x 9 + 37.3491 x 8 + 323.336 x 7 + 2349.42 x 6 + 14355.1 x 5 + 72421.8 x 4 + 290980. x 3 + 874881. x 2 {\displaystyle y'_{n}=0.2167x^{1}0+3.474x^{9}+37.3491x^{8}+323.336x^{7}+2349.42x^{6}+14355.1x^{5}+72421.8x^{4}+290980.x^{3}+874881.x^{2}\!}
+ 1.7517 x 10 6 x + 1.75267 x 10 6 + 2 C 1 e 2 x + C 2 e x {\displaystyle +1.7517x10^{6}x+1.75267x10^{6}+2C_{1}e^{2x}+C_{2}e^{x}\!}
Evaluating at the initial conditions:y ( − 0.75 ) = 828254 + 0.2231301601 C ! + 0.4723665527 C 2 = 1 {\displaystyle y(-0.75)=828254+0.2231301601C_{!}+0.4723665527C_{2}=1\!} y ′ ( − 0.75 ) = 828145 + 0.4462603203 C 1 + 0.4723665527 C 2 = 0 {\displaystyle y'(-0.75)=828145+0.4462603203C_{1}+0.4723665527C_{2}=0\!}
We obtain:C 1 = − 484.022 {\displaystyle C_{1}=-484.022\!} C 2 = − 1753750 {\displaystyle C_{2}=-1753750\!}
Finallyy n = 1753158.594 + 1752673.419 x + 875851.535 x 2 + 291627.134 x 3 + 72745.1129 x 4 + 14484.362 x 5 + 2392.510 x 6 + 335.632 x 7 + 40.417 x 8 {\displaystyle y_{n}=1753158.594+1752673.419x+875851.535x^{2}+291627.134x^{3}+72745.1129x^{4}+14484.362x^{5}+2392.510x^{6}+335.632x^{7}+40.417x^{8}\!} + 4.1499 x 9 + 0.3474 x ( 10 ) + 0.0197 x ( 11 ) − 484.022 e 2 x − 1753750 e x {\displaystyle +4.1499x^{9}+0.3474x^{(}10)+0.0197x^{(}11)-484.022e^{2x}-1753750e^{x}\!}
Here is the graph for this problem using Matlab: